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OverLord2011 [107]

2 years ago

6

Which diagram matches the equation or description.

1 answer:

marissa [1.9K]2 years ago

4 0

I believe the answer to this diagram should be <u><em>Diagram B</em></u>

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Dimas [21]

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I need help in khan academy ASAP!!

Anika [276]

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2 years ago

Line m and p are perpendicular. if the slope of line m is -3, what is the slope of line p

telo118 [61]

Perpendicular slopes multiply to -1, so are <em>negative reciprocals.</em>

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3 years ago

A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr

seraphim [82]

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

$Perimeter = a\sqrt{2} + b\sqrt{10}$

Required

$a + b$

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$(x_1,y_1) = A(0,1)$

$(x_2,y_2) = B(3,4)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}$

$AB = \sqrt{( - 3)^2 + (-3)^2}$

$AB = \sqrt{9+ 9}$

$AB = \sqrt{18}$

$AB = \sqrt{9*2}$

$AB = \sqrt{9}*\sqrt{2}$

$AB = 3\sqrt{2}$

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

$(x_1,y_1) = B (3,4)$

$(x_2,y_2) = C(4,3)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}$

$BC = \sqrt{(-1)^2 + (1)^2}$

$BC = \sqrt{1 + 1}$

$BC = \sqrt{2}$

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = C(4,3)$

$(x_2,y_2) = D (3,0)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}$

$CD = \sqrt{(1)^2 + (3)^2}$

$CD = \sqrt{1 + 9}$

$CD = \sqrt{10}$

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = D (3,0)$

$(x_2,y_2) = A (0,1)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}$

$DA = \sqrt{(3)^2 + (- 1)^2}$

$DA = \sqrt{9 + 1}$

$DA = \sqrt{10}$

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

$Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}$

$Perimeter = 4\sqrt{2} + 2\sqrt{10}$

Recall that

$Perimeter = a\sqrt{2} + b\sqrt{10}$

This implies that

$a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}$

By comparison

$a\sqrt{2} = 4\sqrt{2}$

Divide both sides by $\sqrt{2}$

$a = 4$

By comparison

$b\sqrt{10} = 2\sqrt{10}$

Divide both sides by $\sqrt{10}$

$b = 2$

Hence,

a + b = 2 + 10

a + b = 12

3 0

2 years ago

What is the solution set for 5x>15?

denis23 [38]

Answer= X>3

5x>15

Divide both sides by 5x

$\frac{5 x}{5x} > \frac{15}{5x}$

5x will cancel with 15. You get 3

So, x>3

6 0

3 years ago